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A134191
Impure numbers in the Collatz (3x+1) iteration.
2
2, 4, 5, 8, 10, 11, 13, 14, 16, 17, 20, 22, 23, 26, 28, 29, 31, 32, 34, 35, 38, 40, 41, 44, 46, 47, 49, 50, 52, 53, 56, 58, 59, 61, 62, 64, 65, 67, 68, 70, 71, 74, 76, 77, 80, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 98, 100, 101, 103, 104, 106, 107, 110, 112, 113, 116, 118
OFFSET
1,1
COMMENTS
Let f(k) be the trajectory of the Collatz iteration of the number k. Then Shaw calls a number n impure if n is in f(k) for some k < n. Shaw has an algorithm for finding congruences that the impure numbers satisfy.
LINKS
Douglas J. Shaw, The Pure Numbers Generated by the Collatz Sequence, The Fibonacci Quarterly, Vol. 44, Number 3, August 2006, pp. 194-201.
FORMULA
Complement of A061641.
EXAMPLE
The Collatz trajectory of 3 is (3,10,5,16,8,4,2,1), showing that the numbers 4,5,8,10,16 are impure.
MATHEMATICA
c[n_] := If[EvenQ[n], n/2, 3n + 1]; nn=1000; t=Table[0, {nn}]; Do[If[t[[n]]==0, m=n; While[m=c[m]; If[nn>=m>n && t[[m]]==0, t[[m]]=n]; m>nn || t[[m]]>0]], {n, nn}]; Flatten[Position[t, _?(#>0&)]]
CROSSREFS
Sequence in context: A094591 A189093 A325442 * A286803 A026138 A026170
KEYWORD
nonn
AUTHOR
T. D. Noe, Oct 12 2007
STATUS
approved